Tangent Line Slope: Polar Curve R = 2 - 6sin(θ) At (8, 3π/2)

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Let's dive into the fascinating world of polar curves and explore how to determine the slope of a tangent line at a specific point. In this comprehensive guide, we'll tackle the challenge of finding the slope of the tangent line to the polar curve r = 2 - 6sin(θ) at the point (8, 3π/2). This topic falls squarely within the realm of calculus, specifically dealing with polar coordinates and their applications. Understanding this process is crucial for anyone studying calculus, engineering, or related fields. So, buckle up, and let's get started!

Understanding Polar Coordinates and Curves

Before we jump into the calculations, let’s establish a firm understanding of polar coordinates and polar curves. Unlike the familiar Cartesian coordinate system (x, y), polar coordinates represent a point using a distance r from the origin (or pole) and an angle θ from the positive x-axis. This gives us a coordinate pair (r, θ). Polar curves are then defined by equations that relate r and θ, creating a variety of interesting shapes that are often difficult to represent in Cartesian coordinates. Visualizing these curves is the first step to truly understanding them.

The equation r = 2 - 6sin(θ) represents a specific type of polar curve. The sine function introduces a vertical element, causing the curve to have a shape that's stretched or compressed along the y-axis. The constants 2 and -6 influence the size and position of the curve relative to the pole. To fully grasp the curve's behavior, imagine plotting points for different values of θ. You'll notice that as θ changes, r varies, tracing out the curve's unique form. This curve is a limacon, a looped shape that's symmetric about the y-axis. Polar coordinates provide an elegant way to express such shapes, which might require more complex equations in Cartesian form.

To further solidify your understanding, consider the point (8, 3π/2). This point is given in polar coordinates. The angle 3π/2 corresponds to the negative y-axis, and the distance r = 8 indicates that the point is 8 units away from the origin along that direction. It's crucial to recognize that polar coordinates are not unique; the same point can be represented by multiple coordinate pairs. For example, adding 2π to θ will result in the same point. Similarly, using a negative r and adjusting the angle appropriately can also represent the same point. This non-uniqueness is a characteristic feature of polar coordinates and needs to be kept in mind when working with polar curves.

Converting to Cartesian Coordinates

To find the slope of the tangent line, we'll need to transition from polar coordinates (r, θ) to Cartesian coordinates (x, y). The fundamental relationships that connect these two systems are:

  • x = r cos(θ)
  • y = r sin(θ)

These equations allow us to express any point in polar coordinates as a point in Cartesian coordinates and vice versa. They are derived from basic trigonometry, considering the right triangle formed by the point (x, y), the origin, and the projection of the point onto the x-axis. The distance r is the hypotenuse of this triangle, and x and y are the adjacent and opposite sides, respectively, with respect to the angle θ. Mastering these conversions is essential for dealing with problems that involve both polar and Cartesian coordinate systems.

Given our polar equation r = 2 - 6sin(θ), we can substitute this expression for r into the conversion equations. This yields:

  • x = (2 - 6sin(θ)) cos(θ) = 2cos(θ) - 6sin(θ)cos(θ)
  • y = (2 - 6sin(θ)) sin(θ) = 2sin(θ) - 6sin²(θ)

Now, we have x and y expressed as functions of θ. This is a crucial step because it allows us to use the chain rule of calculus to find dy/dx, which represents the slope of the tangent line in Cartesian coordinates. By expressing x and y in terms of the parameter θ, we've effectively parameterized the curve. This technique is widely used in calculus to analyze curves that are not easily expressed as functions of x or y alone. Understanding this conversion process is key to unlocking the power of polar calculus.

Applying Calculus: Finding dy/dθ and dx/dθ

The next step involves the application of calculus to find the derivatives dy/dθ and dx/dθ. These derivatives represent the rate of change of y and x with respect to the angle θ. To compute these, we'll differentiate the expressions we derived earlier:

  • x(θ) = 2cos(θ) - 6sin(θ)cos(θ)
  • y(θ) = 2sin(θ) - 6sin²(θ)

Using the product rule and chain rule where necessary, we get:

  • dx/dθ = -2sin(θ) - 6[cos(θ)cos(θ) + sin(θ)(-sin(θ))] = -2sin(θ) - 6cos²(θ) + 6sin²(θ)
  • dy/dθ = 2cos(θ) - 12sin(θ)cos(θ)

These derivatives are fundamental to finding the slope of the tangent line. The derivative dx/dθ tells us how the x-coordinate changes as the angle θ changes, and dy/dθ tells us the same for the y-coordinate. These rates of change are crucial for understanding the curve's behavior at any given point. For instance, if dy/dθ is positive, the y-coordinate is increasing as θ increases, and vice versa. Similarly, the sign of dx/dθ indicates whether the x-coordinate is increasing or decreasing.

The process of differentiation requires a solid understanding of basic trigonometric derivatives, the product rule, and the chain rule. The product rule is essential when differentiating terms like sin(θ)cos(θ), and the chain rule is needed when dealing with composite functions like sin²(θ). Mastering these differentiation techniques is not just important for this problem but is a cornerstone of calculus in general. Practicing these techniques on various functions will build your proficiency and confidence in calculus.

Calculating the Slope: dy/dx

Now, we can calculate the slope of the tangent line, dy/dx, using the chain rule. The chain rule provides a way to find the derivative of a composite function, and in this case, it allows us to relate dy/dx to dy/dθ and dx/dθ. The relationship is given by:

  • dy/dx = (dy/dθ) / (dx/dθ)

This formula is a direct consequence of the chain rule and is a powerful tool for finding derivatives in parametric equations and polar coordinates. It allows us to express the slope of the tangent line in terms of θ, which is exactly what we need given our expressions for dy/dθ and dx/dθ.

Plugging in the expressions we found earlier, we have:

  • dy/dx = (2cos(θ) - 12sin(θ)cos(θ)) / (-2sin(θ) - 6cos²(θ) + 6sin²(θ))

This expression gives us the slope of the tangent line at any point on the polar curve, as long as we know the corresponding angle θ. It's a powerful result that encapsulates the relationship between the rates of change of x and y with respect to θ. To find the slope at a specific point, we simply need to evaluate this expression at the θ value corresponding to that point.

Evaluating at the Point (8, 3π/2)

To find the slope of the tangent line at the point (8, 3π/2), we substitute θ = 3π/2 into our expression for dy/dx:

  • dy/dx = (2cos(3π/2) - 12sin(3π/2)cos(3π/2)) / (-2sin(3π/2) - 6cos²(3π/2) + 6sin²(3π/2))

Now, we evaluate the trigonometric functions at θ = 3π/2:

  • cos(3π/2) = 0
  • sin(3π/2) = -1

Substituting these values into the expression for dy/dx, we get:

  • dy/dx = (2(0) - 12(-1)(0)) / (-2(-1) - 6(0)² + 6(-1)²) = 0 / (2 + 0 + 6) = 0 / 8 = 0

Therefore, the slope of the tangent line to the polar curve r = 2 - 6sin(θ) at the point (8, 3π/2) is 0. This result tells us that the tangent line at this point is horizontal. A slope of 0 corresponds to a horizontal line because the y-coordinate is not changing with respect to the x-coordinate at this point. Understanding the geometric interpretation of the slope is crucial for visualizing the behavior of the curve and its tangent lines.

Conclusion

In conclusion, we've successfully navigated the process of finding the slope of the tangent line to the polar curve r = 2 - 6sin(θ) at the point (8, 3π/2). We started by understanding polar coordinates and curves, then converted the polar equation to Cartesian form, applied calculus to find dy/dθ and dx/dθ, and finally, calculated the slope dy/dx. By substituting θ = 3π/2, we found that the slope of the tangent line at the given point is 0, indicating a horizontal tangent line.

This problem demonstrates the power of calculus in analyzing curves defined in polar coordinates. The techniques we've used, such as converting between polar and Cartesian coordinates, applying the chain rule, and evaluating trigonometric functions, are fundamental to calculus and have wide applications in various fields. Mastering these techniques will significantly enhance your problem-solving abilities in calculus and related disciplines.

For further exploration and a deeper understanding of polar curves and calculus, visit Khan Academy's Calculus section on Polar Coordinates. This resource provides a wealth of information, examples, and exercises to help you master this fascinating topic.